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Question

If p1,p2,p3 denote the distance of the plane 2x3y+4z+2=0 from the planes 2x3y+4z+6=0,4x6y+8z+3=0 and 2x3y+4z6=0 respectively, then

A
p1+8p2p3=0
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B
p32=16p22
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C
8p22=p12
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D
p1+2p2+3p3=29
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Solution

The correct option is C p1+8p2p3=0

Since the planes are all parallel planes,

p1=|26|22+32+42=44+9+16=429

Equation of the plane 4x6y+8z+3=0 can be written as 2x3y+4z+32=0

So, p2=23222+32+42=1229

and p3=|2+6|22+32+42=829

p1+8p2p3=0


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